022: Fifth Powers (8, 10, 12 terms)

This is a generalization of a single numerical result by Tarry. It can be the case that a k = 1,3,5 can give rise to a k = 2,4,6. Given,

(3p+q)^k + (p+q)^k + (p+r)^k + (-p+r)^k + (-3p+q)^k + (-p+q)^k + u^k + v^k = 0, for k = 1,3,5, then,

(4p+q)^k + (2p+r)^k + (p+u)^k + (p+v)^k = (4p-q)^k + (2p-r)^k + (p-u)^k + (p-v)^k, for k = 2,4,6

which has a complete soln in terms of an elliptic curve as,

{p,q,r,u,v} = {2x²+x+1, x²+6x+1, -8(x²+2x), 6x²+4x-2+y, 6x²+4x-2-y}

and where x,y must satisfy,

y² = 15x⁴+104x³+223x²+54x+4

though some of the small rational solns are trivial. (Update, 8/4/09): Alternatively, the system can be expressed as,

(a-2b+c)^k + (a-2b-c)^k + (b+3c)^k + (b+c)^k = (a-d)^k + (a+d)^k + (-b+3c)^k + (-b+c)^k, for k = 1,3,5, then,

(a+c-d)^k + (a+c+d)^k + (-a+2b+2c)^k + (b-4c)^k = (a-c-d)^k + (a-c+d)^k + (-a+2b-2c)^k + (b+4c)^k, for k = 1,2,4,6,

where {a,b,c,d} satisfy a²+ab+2b² = 8c² and a²-7ab+42b² = 8d².

This is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-1/3, -1, 2, 5} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {15,1}. (End update)

III. Constraint: x₁-x₂ = x₃-x₄ = -(x₅-x₆)

(p+aq)^k + (p+bq)^k + (p+cq)^k + (d+q)^k = (p-aq)^k + (p-bq)^k + (p-cq)^k + (d-q)^k, for k = 1,3,5

already satisfying the constraint (after appropriate transposition of terms). Solving this system involves only making a quartic and quadratic polynomial into squares, hence is a slight improvement over the previous method. (End update.)

(a+m)^k + (a-m)^k + (b+m)^k + (b-m)^k + (c+m)^k + (c-m)^k = (d+n)^k + (d-n)^k

Expanding for k = 1,3,5, we can eliminate d,n and find the resultant in terms of m which is only a quadratic. Explicitly, the complete soln is, d = a+b+c, and {m,n} = {u/(2d), v/(2d)} where {u,v} are,

(a+b)⁴ + (a+b)(4a²+9ab+4b²)c + (3a+2b)(2a+3b)c² + 4(a+b)c³ + c⁴ = u²

(a²-b²)² – 3ab(a+b)c – (2a²+3ab+2b²)c² + c⁴ = v²

One must find {a,b,c} such that these two quartic polynomials in c are squares which in general is not easily done. (Surprisingly, for a k = 1,3,5,7 with a similar constraint, it involves only two quadratics to be made squares which is easier.) Without loss of generality, we can set a = 1 and one soln for the first eqn is,

c = -(2+3b)/(3+4b)

It reduces the second eqn, after removing square factors, to solving,

25+24b-48b²+64b⁴ = y²

with initial b = 3/16 from which other rational points can then be computed. (Update, 8/13/09): One can use a variant,

II. Constraint: x₁-x₂ = x₃-x₄ = x₅-x₆

We can focus first on a special case. Define the eqn,

(a+c)^k + (a-c)^k + (b+c)^k + (b-c)^k + (a+b+c)^k + (a+b-c)^k + (u)^k + (v)^k = 0, for k = 1,3,5, (eq.1)

This has the common difference of the first three pairs of terms as 2c and also has a complete parametrization, but now involves an elliptic curve. Note that this also has the sum of its first four terms as x₁+x₂+x₃+x₄ = x₅+x₆. Let {a,b,u,v} = {p+q, -p+q, d-4q, -d-4q}, then eq.1 is true if,

c²+11d² = 4p², and 3c²+d² = 12q²,

where the ratio c/d = {1, 1/3} should be avoided. A small soln is {c,d} = {37, 15} and it is easy to convert the two quadratic conditions into an elliptic curve. A special case solvable as quadratic forms is,

(a+b+c)^k + (a+b-c)^k + (-a+b+c)^k + (-a+b-c)^k + (8b+c)^k + (8b-c)^k = (7b)^k + (13b)^k, for k = 1,3,5, (eq.2)

where -a²+126b² = 5c².

The common difference of the first three pairs is also 2c but the sum of its first four terms is now x₁+x₂+x₃+x₄ = (x₅+x₆)/4. From the initial soln {a,b,c} = {1,1,5}, one can get a parametrization. However, for the complete soln to the general case, let,

Piezas

So far, most of the identities discussed have the condition x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0. One can avoid this and explore other constraints such as:

a) x₁+x₂ = x₃+x₄ = x₅+x₆, as (a+c)^k + (-a+c)^k + (b+c)^k + (-b+c)^k + (a+b+c)^k + (-a-b+c)^k + u^k + v^k = 0

b) x₁-x₂ = x₃-x₄ = x₅-x₆, as (a+c)^k + (a-c)^k + (b+c)^k + (b-c)^k + (a+b+c)^k + (a+b-c)^k + u^k + v^k = 0

c) x₁-x₂ = x₃-x₄ = -(x₅-x₆), as (3p+q)^k + (p+q)^k + (p+r)^k + (-p+r)^k + (-3p+q)^k + (-p+q)^k + u^k + v^k = 0

I. Constraint: x₁+x₂ = x₃+x₄ = x₅+x₆

We can focus on a special case of this constraint by defining the polynomial, P(k): = (a+c)^k + (-a+c)^k + (b+c)^k +(-b+c)^k + (a+b+c)^k + (-a-b+c)^k which is one side of the Kawada-Wooley equation. This has the common sum of two terms as 2c. Note this also obeys x₁-x₂+x₃-x₄ = x₅-x₆. If we set the equation,

E(k): = P(k) + (r)^k + (s)^k = 0

to be simultaneously true for k = 1,3,5, then this has a complete parametrization. By eliminating {r,s} from the three equations E(1), E(3), E(5) using resultants, easily done using Mathematica’s Resultant[] function, one gets a final eqn,

(a²+ab+b²-28c²) (a²+ab+b²-4c²) = 0

so there are two identities involving P(k), namely,

(a+c)^k + (-a+c)^k + (b+c)^k +(-b+c)^k + (a+b+c)^k + (-a-b+c)^k = 2(3c)^k, where a²+ab+b² = 4c², and,

(a+c)^k + (-a+c)^k + (b+c)^k +(-b+c)^k + (a+b+c)^k + (-a-b+c)^k = (-c)^k + (7c)^k, where a²+ab+b² = 28c²,

both for k = 1,3,5. Also, the polynomial P_k as a sum is quite peculiar since a common factor turns up for k = 1,2,3,4,5. To illustrate, for the first, set {a,b,c} = {2(u²-v²), 2(2uv+v²), (u²+uv+v²)}, then for k = 1,2,3,4,5 we have the sums,

P_k = {2(3)(u²+uv+v²), 2(11)(u²+uv+v²)², 2(3³)(u²+uv+v²)³, 2(83)(u²+uv+v²)⁴, 2(3⁵)(u²+uv+v²)⁵}

while for the second, let, {a,b,c} = {2(u²+6uv+2v²), 2(2u²-2uv-3v²), (u²+uv+v²)}, then,

P_k + c^k = {7(u²+uv+v²), 7(17)(u²+uv+v²)², 7³(u²+uv+v²)³, 7(545)(u²+uv+v²)⁴, 7⁵(u²+uv+v²)⁵}

Thus, in general, for these k and no higher, they have the common factor u²+uv+v². These identities also have other properties. The first, by solving a²+ab+b² = t^2k, gives a soln to,

x₁⁵ + x₂⁵ + x₃⁵ + x₄⁵ + x₅⁵ + x₆⁵ = 2(3⁵y^5k)

the quintic analogue of Ramanujan’s family of solns to x₁⁴ + x₂⁴ + x₃⁴ = 2(y^2k). See Form 11 of Fourth Powers. Also, since the a,b,c are binary quadratic forms, these can provide solns to,

x₁⁵ + x₂⁵ + x₃⁵ + x₄⁵ + x₅⁵ + x₆⁵ + x₇⁵ = 1

by setting one term as equal to ±1 and solving the resulting Pell eqn. For ex, we have,

(a+c)^k + (b+c)^k + (a+b+c)^k + (-a-b+c)^k + (-a+c)^k + 2(-3c)^k = ±1

where {a,b,c} = {2(p+q)(p+5q), 8(p+4q)q, p²+8pq+19q²}, for k = 1,3,5, if p²-13q² = ±1.

(x+z)⁵ + (-x+z)⁵ + (y+z)⁵ + (-y+z)⁵ + (x+y+z)⁵ + (-x-y+z)⁵ = 66z⁵

where x²+xy+y² = z², call this condition C₁, the smallest soln of which is {x,y,z} = {3,5,7}. If we choose to express 66 as a sum of powers 2(1)⁵+2(2)⁵, this results in a 10-term identity,

(x+z)^k + (-x+z)^k + (y+z)^k + (-y+z)^k + (x+y+z)^k + (-x-y+z)^k = 2(z)^k + 2(2z)^k,

with the same conditon C₁, though it is now valid for k = 1,2,3,4,5. This is also discussed in Form 5.5 below. But if we let z² = nh², with n = 4 or 28, then the KW identity is,

(h+x)⁵ + (h-x)⁵ + (h+y)⁵ + (h-y)⁵ + (h+x+y)⁵ + (h-x-y)⁵ = mh⁵

for m = {486, 16806}, where x²+xy+y² = nh² for n = {4, 28}, respectively. But if we express m as a sum of powers, then,

(h+x)^k + (h-x)^k + (h+y)^k + (h-y)^k + (h+x+y)^k + (h-x-y)^k = 2(3h)^k, where x²+xy+y² = 4h²

(h+x)^k + (h-x)^k + (h+y)^k + (h-y)^k + (h+x+y)^k + (h-x-y)^k = (-h)^k + (7h)^k, where x²+xy+y² = 28h²

now valid for k = 1,3,5. These will be discussed in Constraint I immediately after this note, though derived differently using resultants. (End update.)

One can test it with a random example, say, {a,b,c,d} = {21, -13, 9, -17} and {e,f,g,h} = {23, -3, -21, 1}, though it will work in the general case. Note that a+b+c+d = e+f+g+h for this system implies it is also valid for k = 2. [6] has already been given in Fourth Powers but is related to the Chernick-Lander theorem given in the previous section and is analogous to Ramanujan’s 6-10-8 Identity. [3] seems to have analogous versions for any system k = 1,2,3,5,…n for consecutive odd exponents n > 3, while [1] is a general identity that works on four terms a+b+c+d = 0.

Proof: Given x₁^k+x₂^k+x₃^k+x₄^k - (y₁^k+y₂^k+y₃^k+y₄^k) = 0 such that x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0, this is completely parametricized by the Chernick-Lander form, call this L(k),

L(k): = (a+b+c)^k + (a-b-c)^k + (-a-b+c)^k + (-a+b-c)^k – ((d+e+f)^k + (d-e-f)^k + (-d-e+f)^k + (-d+e-f)^k)

since the three variables {a,b,c} can suffice to express the x_i (and likewise {d,e,f} for the y_i). Expanding L(k) for k = 1,2,3,5, we get,

L(1) = 0, L(2):= a²+b²+c²-d²-e²-f² ,

L(3): = abc-def, L(5): = abc(a²+b²+c²)-def(d²-e²-f²),

If we wish to set all L(k) = 0, if L(3) = 0, one can see that the eqn L(5) = 0 reduces to just L(2) = 0. Eliminating f between L(2) = 0 and L(3) = abc-def = 0, we get a polynomial, call it P. Substituting the terms of L(k) on any of the statements of Theorem 1 after [1], we get an expression in {a,b,c,d,e,f} and, eliminating f between this and abc-def = 0, we get P again (as a factor). Thus, the roots of P = 0 satisfy both the condition and the theorem, which proves the latter. (For [1], simply substract one side from the other, expand the identity, and one can see it is true if a+b+c+d = 0.)

Lander

(a+b+c)⁵ + (a-b-c)⁵ + (-a-b+c)⁵ + (-a+b-c)⁵ = 80abc(a²+b²+c²)

(Update, 12/13/09): Tony Rizzo showed that there are non-trivial solns to,

80abc(a²+b²+c²) = d⁵+e⁵ (eq.1)

thus yielding a (5.3.3) and gave the example {a,b,c,d,e} = {32, 25, 50, 80, 100}. More generally, this gives a (5.3.3) with the side condition x₁+x₂ = y₁+y₂. The Sastry-Chowla identity has this constraint, so there is in fact an infinite number of solns to eq.1 given by {a,b,c,d,e} = {u⁵, 25v⁵, 50v⁵, 10u³v², 50uv⁴}, for arbitrary {u,v}, with Rizzo's as {u,v} = {2,1}. However, there are other solns as the Sastry-Chowla identity cannot give a (5.3.3) with all terms positive. Duncan Moore has a database here and out of around 5400 solns, only 12 had the required condition that x₁+x₂ = y₁+y₂. Q: Is there any other family that solves eq.1? (End update.)

Lander (k = 1,5)

(ax²-vxy+w₁y²)^k + (ax²+vxy+w₂y²)^k + (x²-vxy+w₃y²)^k + (x²+vxy+w₄y²)^k =

(ax²+vxy+w₁y²)^k + (ax²-vxy+w₂y²)^k + (x²+vxy+w₃y²)^k + (x²-vxy+w₄y²)^k

This has the side condition x₁+x₂ = y₁+y₂; x₃+x₄ = y₃+y₄ which for odd powers is equivalent to x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0 and was discussed in the previous section. However, this new one is only for k = 1,5 but has a surprising property. Let v = 3a², and {w₁, w₂, w₃, w₄} = {p+q, p-q, r+s, r-s}. Then, the soln is,

{p,q,r,s} = {a⁷+a⁵-2a³+a, 3a², a⁶-2a⁴+a²+1, 3a⁵}

Note that {p,q,r,s} also solve,

(p+q)⁴ + (r-s)⁴ = (p-q)⁴ + (r+s)⁴

To see this, if the 8-term eqn above was expanded and powers of {x,y} collected, we get a system of four auxiliary eqns in four unknowns, one of which is,

w₁⁴-w₂⁴- w₃⁴+w₄⁴ = 0

explaining why it has this property. In fact, by solving this system one can directly derive the w_i, instead of the geometric derivation which Lander used. (Note: Are there other forms of (5,4,4) that also involve the other parametrizations of (4,2,2)?)

Kawada, Wooley

(h+x)⁵ + (h-x)⁵ + (h+y)⁵ + (h-y)⁵ + (h+x+y)⁵ + (h-x-y)⁵ = 20h(x²+xy+y²+h²)² - 14h⁵

(Update, 11/2/09): Tony Rizzo pointed out we can let x²+xy+y² = z² for some z. He gave Ramanujan's simple soln to this as {x,y} = {a-d, b+c+d} where a/b = c/d, and b = c. For the special case when y = -x, this then becomes,

2(h+x)⁵ + 2(h-x)⁵ + 2h⁵ = 20h(h²+x²)² - 14h⁵

where the whole eqn is now div by 2. Note: I realized Rizzo's observation can be extended to cover some of the identities I found given below. For example, let z = h, and the Kawada-Wooley (KW) identity becomes,

Combining [2] and [3] yields the same relation as combining [4] and [5], namely,

[6] 7(a⁴+b⁴+c⁴+d⁴-e⁴-f⁴-g⁴-h⁴)(a⁹+b⁹+c⁹+d⁹-e⁹-f⁹-g⁹-h⁹) = 12(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶)(a⁷+b⁷+c⁷+d⁷-e⁷-f⁷-g⁷-h⁷), and,

4. x₁+x₂ = y₁+y₂; x₃+x₄ = y₃+y₄

Sixth powers:

1. x₁+x₂ = y₁+y₂; x₃+x₄ = y₃+y₄

Seventh powers:

1. x₁+x₂ = x₃+x₄ = y₁+y₂

2. x₁-x₂ = x₃-x₄ = y₁-y₂

3. x₁-x₂ = y₁-y₂; x₃-x₄ = y₃-y₄

Eighth powers:

1. x₁-x₂ = y₁-y₂; x₃-x₄ = y₃-y₄

Of course these are not all of the possible side conditions but small solns tend to follow them hence are the first ones noticed. They are useful since they give more “handles” to deal with the system and, in some cases, reduce it to merely solving a quadratic or making a low-degree polynomial a square. (It is tempting to speculate ninth and tenth powers may obey similar rules though, since only one soln is known for each, very little can be said about their behavior.) (Update: More solns have been found for tenth powers. See update in last section.) Side conditions can also have interesting implications,

Piezas

Theorem 1: If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k, for k = 1,3,5 where a+b+c+d = e+f+g+h = 0, then the following six relations are also true,

[1] 5(a²+b²+c²+d²)(a³+b³+c³+d³) = 6(a⁵+b⁵+c⁵+d⁵)

[2] 3(a²+b²+c²+d²)(a⁴+b⁴+c⁴+d⁴-e⁴-f⁴-g⁴-h⁴) = 4(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶)

[3] 9(a²+b²+c²+d²)(a⁷+b⁷+c⁷+d⁷-e⁷-f⁷-g⁷-h⁷) = 7(a⁹+b⁹+c⁹+d⁹-e⁹-f⁹-g⁹-h⁹)

[4] (a³+b³+c³+d³)(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶) = (a⁹+b⁹+c⁹+d⁹-e⁹-f⁹-g⁹-h⁹)

[5] 7(a³+b³+c³+d³)(a⁴+b⁴+c⁴+d⁴-e⁴-f⁴-g⁴-h⁴) = 12(a⁷+b⁷+c⁷+d⁷-e⁷-f⁷-g⁷-h⁷)